NivaarExam PrepOfficial exam papers ↗

18-Env-A2 Hydrology and Municipal Hydraulics Engineering · May 2014

Question 7 of 7: Open-Channel Flow and Sediment Transport

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2014 — 04-Env-A2 / Hydrology and Municipal Hydraulics Engineering. 3 hours duration; closed book with an 8.5×11 in double-sided aid sheet; Casio or Sharp approved calculator only. Any five questions constitute a complete paper (first five answers marked); all seven are solved below for completeness. Each question is worth 20 marks.

Reference texts. Chow, Open-Channel Hydraulics; Linsley, Kohler & Paulhus, Hydrology for Engineers (3rd ed.); Walski et al., Advanced Water Distribution Modeling and Management; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.).

Question 7: Open-Channel Flow and Sediment Transport (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(i) Trapezoidal Channel — Discharge and Reynolds Number

Given. Grass-lined trapezoidal channel, uniform flow:

Given data
QuantitySymbolValue
Normal depth$y$5 m
Base width$b$10 m
Side slope, H:V$z$ (=H/V)1:5 → $z=0.2$
Bed slope$S_o$4% = 0.04
Manning's roughness (assumed, grass lining)$n$0.035
b = 10 m y = 5 m side slope H:V = 1:5 (z = 0.2) Grass-lined trapezoidal channel, normal depth
Trapezoidal channel cross-section used for the uniform-flow calculation.
Check: (1) Manning's $n=0.035$ is assumed for a grass lining (not stated in the question) — a typical "good stand, moderate retardance" design value. (2) The side-slope ratio is used exactly as printed, H:V $=1:5$ ($z=0.2$, i.e. quite steep, 1 m horizontal per 5 m vertical); this is atypical for a mowable grass lining (which normally uses a much flatter 3:1 to 5:1 H:V slope for stability and maintenance access), and it is the main reason the resulting velocity below is far higher than a real grass channel could sustain without eroding — the arithmetic is carried through exactly as the stated geometry and slope require.

Find. The discharge $Q$ and the Reynolds number/flow regime.

Approach. Compute the trapezoidal section properties, apply Manning's Equation for $Q$, then evaluate the open-channel Reynolds number $Re=VR/\nu$.

  1. Section properties. $$A = (b+zy)y = (10+0.2\times5)(5) = \boxed{55.0\ \text{m}^2}$$ $$P = b + 2y\sqrt{1+z^2} = 10 + 2(5)\sqrt{1+0.2^2} = \boxed{20.20\ \text{m}}, \qquad R = \frac{A}{P} = \frac{55.0}{20.20} = \boxed{2.72\ \text{m}}.$$
  2. (a) Manning's Equation. $$Q = \frac{1}{n}AR^{2/3}S_o^{1/2} = \frac{1}{0.035}(55.0)(2.72)^{2/3}(0.04)^{1/2} = \boxed{613\ \text{m}^3/\text{s}}.$$
  3. (b) Reynolds number. $V = Q/A = 613/55.0 = 11.14$ m/s; using the open-channel convention $Re=VR/\nu$: $$Re = \frac{(11.14)(2.72)}{1.0\times10^{-6}} = \boxed{3.0\times10^{7}}.$$ Since $Re \gg 12{,}500$, the flow is turbulent (as is essentially unavoidable at this depth, slope and velocity).
QuantityValue
Flow area, $A$55.0 m²
Hydraulic radius, $R$2.72 m
Discharge, $Q$≈ 613 m³/s
Velocity, $V$11.1 m/s
Reynolds number, $Re$$3.0\times10^{7}$ (turbulent)

(ii) Specific Energy Across a Bed Rise

Given. Same trapezoidal channel ($b=10$ m, $z=0.2$); flow $Q=30$ m³/s; upstream normal depth $Y_1=4$ m; bed rises $\Delta Z=1.5$ m over a 7 m reach; frictional losses negligible:

water surface Y₁ Section 1 Y₂ Section 2 (crest) ΔZ = 1.5 m Flow →
Longitudinal profile through the bed rise (bump); specific energy is conserved between sections 1 and 2 since friction is neglected and the datum shifts up by $\Delta Z$.

Find. The flow depth $Y_2$ at the crest of the bump.

Approach. Compute the upstream specific energy $E_1$ and Froude number (to confirm subcritical flow and that the bump does not choke it), apply $E_1 = E_2 + \Delta Z$, then solve the resulting cubic-type specific-energy equation for $Y_2$ numerically.

  1. Section 1 properties and specific energy. $$A_1 = (10+0.2\times4)(4) = 43.2\ \text{m}^2, \qquad V_1 = \frac{30}{43.2} = 0.694\ \text{m/s},$$ $$E_1 = Y_1 + \frac{V_1^2}{2g} = 4 + \frac{(0.694)^2}{2(9.81)} = \boxed{4.025\ \text{m}}.$$
  2. Confirm subcritical flow (no choking). Top width $T_1=b+2zY_1=11.6$ m, hydraulic depth $D_1=A_1/T_1=3.72$ m: $$Fr_1 = \frac{V_1}{\sqrt{gD_1}} = \frac{0.694}{\sqrt{(9.81)(3.72)}} = \boxed{0.115} \; (<1,\ \text{subcritical}).$$ The critical energy for this section shape is only $\approx1.44$ m, well below $E_1-\Delta Z$, so the 1.5 m bump does not choke the flow to critical depth.
  3. Apply energy conservation across the bump. With friction neglected, $E_1 = E_2 + \Delta Z$ (energy measured from each section's own bed): $$E_2 = E_1 - \Delta Z = 4.025 - 1.5 = \boxed{2.525\ \text{m}}.$$
  4. Solve for $Y_2$. Since the approach flow is subcritical and the bump does not choke the flow, the physically valid root is the subcritical branch of $$E_2 = Y_2 + \frac{Q^2}{2g\left[(b+zY_2)Y_2\right]^2},$$ solved numerically (Newton iteration) for $Y_2$: $$\boxed{Y_2 \approx 2.46\ \text{m}}.$$
QuantityValue
Upstream specific energy, $E_1$4.025 m
Upstream Froude number, $Fr_1$0.115 (subcritical)
Specific energy at crest, $E_2$2.525 m
Depth at crest, $Y_2$≈ 2.46 m

(iii) Force Diagram for Incipient Sediment Motion

W Fₑ Fₗ α φ Force balance on a bed particle (incipient motion) flow →
Force balance on a single bed particle on a channel bed inclined at angle α: submerged weight $W$, streamwise drag $F_D$, and bed-normal lift $F_L$.

The diagram resolves the submerged weight $W$ of a bed particle into a component along the bed, $W\sin\alpha$, which (together with the flow's streamwise drag force $F_D$) drives the particle to slide or roll downstream, and a component normal to the bed, $W\cos\alpha$, which provides the normal reaction that generates frictional resistance. The flow's lift force $F_L$ (from the pressure differential/velocity gradient across the particle as flow accelerates over its top) acts normal to the bed in the direction that reduces this normal reaction. The maximum frictional resistance available to hold the particle in place is the normal reaction multiplied by the coefficient of friction $\tan\phi$ (where $\phi$ is the particle's angle of repose): incipient motion occurs at the threshold where the driving forces exactly equal this maximum resistance, $$W\sin\alpha + F_D = \tan\phi\,(W\cos\alpha - F_L),$$ which rearranges directly to the given equation $\tan\phi = \dfrac{W\sin\alpha+F_D}{W\cos\alpha-F_L}$. Physically, this shows the stream needs *less* drag force $F_D$ to initiate motion as either the bed slope $\alpha$ increases (gravity increasingly assists sliding) or as the lift force $F_L$ increases (reducing the normal force and hence the available friction) — which is exactly why steep alluvial channels and high-velocity reaches (both drag and lift grow with velocity) are far more prone to bed and bank erosion than mild, low-velocity channels.

Back to the paper →